87 lines
		
	
	
		
			1.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			87 lines
		
	
	
		
			1.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5a23c84252665b21eecc7edf
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| title: Least common multiple
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| challengeType: 5
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| forumTopicId: 302301
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| ---
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| 
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| # --description--
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| 
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| The least common multiple of 12 and 18 is 36, because 12 is a factor (12 × 3 = 36), and 18 is a factor (18 × 2 = 36), and there is no positive integer less than 36 that has both factors. As a special case, if either *m* or *n* is zero, then the least common multiple is zero. One way to calculate the least common multiple is to iterate all the multiples of *m*, until you find one that is also a multiple of *n*. If you already have *gcd* for [greatest common divisor](<https://rosettacode.org/wiki/greatest common divisor>), then this formula calculates *lcm*. ( \\operatorname{lcm}(m, n) = \\frac{|m \\times n|}{\\operatorname{gcd}(m, n)} )
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| 
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| # --instructions--
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| 
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| Compute the least common multiple of an array of integers. Given *m* and *n*, the least common multiple is the smallest positive integer that has both *m* and *n* as factors.
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| 
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| # --hints--
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| 
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| `LCM` should be a function.
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| 
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| ```js
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| assert(typeof LCM == 'function');
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| ```
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| 
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| `LCM([2, 4, 8])` should return a number.
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| 
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| ```js
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| assert(typeof LCM([2, 4, 8]) == 'number');
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| ```
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| 
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| `LCM([2, 4, 8])` should return `8`.
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| 
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| ```js
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| assert.equal(LCM([2, 4, 8]), 8);
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| ```
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| 
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| `LCM([4, 8, 12])` should return `24`.
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| 
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| ```js
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| assert.equal(LCM([4, 8, 12]), 24);
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| ```
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| 
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| `LCM([3, 4, 5, 12, 40])` should return `120`.
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| 
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| ```js
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| assert.equal(LCM([3, 4, 5, 12, 40]), 120);
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| ```
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| 
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| `LCM([11, 33, 90])` should return `990`.
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| 
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| ```js
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| assert.equal(LCM([11, 33, 90]), 990);
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| ```
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| 
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| `LCM([-50, 25, -45, -18, 90, 447])` should return `67050`.
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| 
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| ```js
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| assert.equal(LCM([-50, 25, -45, -18, 90, 447]), 67050);
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| ```
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| 
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| # --seed--
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| 
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| ## --seed-contents--
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| 
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| ```js
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| function LCM(A) {
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| 
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| }
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| ```
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| 
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| # --solutions--
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| 
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| ```js
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| function LCM(A) {
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|   var n = A.length,
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|     a = Math.abs(A[0]);
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|   for (var i = 1; i < n; i++) {
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|     var b = Math.abs(A[i]),
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|       c = a;
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|     while (a && b) {
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|       a > b ? (a %= b) : (b %= a);
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|     }
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|     a = Math.abs(c * A[i]) / (a + b);
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|   }
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|   return a;
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| }
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| ```
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